Solve Satellite Altitude Above Earth's Surface | 16.7 kN, -1.43e11 J

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To find the altitude of a communication satellite above Earth's surface, gravitational force and potential energy equations can be utilized. The gravitational force acting on the satellite is 16.7 kN, and the gravitational potential energy is -1.43 x 10^11 J. The radius of the Earth is given as 6.38 x 10^6 m. The discussion emphasizes that while the approach using two unknowns (mass of the satellite and height) is valid, there are alternative methods to simplify the calculations. Overall, the participants confirm that the original method is correct and can lead to the desired solution.
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When in orbit, a communication satellite attracts the Earth with a force of 16.7 kN and the earth-satellite gravitational potential energy (relative to zero at infinite separation) is - 1.43*10^11 J. Find the satellite's altitude above the Earth's surface. The radius of the Earth is 6.38*10^6.

OK, I must be making this harder than it needs to be. What I've been trying to do is to use the formulas for gravitational force to get an equation with two unknow variables (Mass of the satellite and height above Earth's surface) And I do the same for gravitational potential energy. Then, since both equations have the same two unknown variables, I solve for one of them and substitute. Is there another way of doing this? Am I doing it copmletely wrong? Please help me! :confused:
 
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You sound like you have the right idea. Give it a shot.
 
Total Energy of a satellite revolving around Earth is given by:

- \frac{GMm}{2r}

BJ


Note:This post has been edited after Older Dan's remarks.
 
Last edited:
Dr.Brain said:
Potential Energy of a satellite revolving around Earth is given by:

- \frac{GMm}{2r}

BJ
There is no 2 in the potential energy. Perhaps you meant the total energy

U = - \frac{GMm}{r}

T = \frac{1}{2}mv^2 = \frac{r}{2} \left( \frac{mv^2}{r} \right) = \frac{r}{2} \left| F_c \right| = \frac{r}{2} \left( \frac{GMm}{r^2} \right) = \frac{GMm}{2r} = -\frac{1}{2} U

E\ \ =\ \ T\ \ +\ \ U = \frac{GMm}{2r}\ \ -\ \ \frac{GMm}{r}\ \ =\ \ - \frac{GMm}{2r}
 
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